cho biểu thức
\(p=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\frac{4\sqrt{x}}{3}\) với x>= 0
a)rút gọn biểu thức
b)tìm x để P=\(\frac{8}{9}\)
c)tìm Max,Min của P
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Nguyễn Thị Anh
2 tháng 8 2016
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7 tháng 8 2016
diều kiện x >= 0
P=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)
= \(\frac{x+2-x+\sqrt{x}-1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)
=\(\frac{\sqrt{x}+1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)=\(\frac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
P=8/9
<=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
<=> \(3\sqrt{x}=2x-2\sqrt{x}+1\)
<=> \(2x-5\sqrt{x}+2=0\)
<=> \(\left[\begin{array}{nghiempt}x=4\\x=\frac{1}{4}\end{array}\right.\)
vậy x=4 hoặc x=1/4 thì p=8/9
7 tháng 8 2016
a) \(P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\left(ĐK:x\ge0;x\ne-1\right)\)
\(=\left[\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Để P=8/9
\(\Leftrightarrow\)\(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)
\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)
\(\Leftrightarrow24x-60\sqrt{x}+24=0\)
\(\Leftrightarrow12\left(2x-5\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x}\right)-\left(4\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)
24 tháng 6 2021
a) đk: x\(\ge0\);
P = \(\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right].\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Để P = \(\dfrac{8}{9}\)
<=> \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
<=> \(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{3}\)
<=> \(\dfrac{3\sqrt{x}-2x+2\sqrt{x}-2}{3\left(x-\sqrt{x}+1\right)}=0\)
<=> \(-2x+5\sqrt{x}-2=0\)
<=> \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
<=> \(\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
c)
Đặt \(\sqrt{x}=a\) (\(a\ge0\))
P = \(\dfrac{4a}{3\left(a^2-a+1\right)}\)
Xét P + \(\dfrac{4}{9}\) = \(\dfrac{4a}{3a^2-3a+3}+\dfrac{4}{9}=\dfrac{12a+4a^2-4a+4}{9\left(a^2-a+1\right)}=\dfrac{4a^2+8a+4}{9\left(a^2-a+1\right)}=\dfrac{4\left(a+1\right)^2}{9\left(a^2-a+1\right)}\ge0\)
Dấu "=" <=> a = -1 (loại)
=> Không tìm được Min của P
Xét P - \(\dfrac{4}{3}\) = \(\dfrac{4a}{3\left(a^2-a+1\right)}-\dfrac{4}{3}=\dfrac{4a-4a^2+4a-4}{3\left(a^2-a+1\right)}=\dfrac{-4a^2+8a-4}{3\left(a^2-a+1\right)}=\dfrac{-4\left(a-1\right)^2}{3\left(a^2-a+1\right)}\le0\)
<=> \(P\le\dfrac{4}{3}\)
Dấu "=" <=> a = 1 <=> x = 1 (tm)
17 tháng 10 2018
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
10 tháng 8 2021
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
10 tháng 8 2021
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)